Duration of a trajectory
Suppose we launch an object straight up. We wish to find how long it will take to return. Suppose we launch it up at a speed \(v_0\). It is well known that the classical escape velocity is given by \[ v_e=\sqrt{\frac{2MG}{R}} \] By examining the energy equation, we find that the speed when the object is a distance r from the center of the planet is given by: \[ v(r)=v_e\sqrt{\frac{R}{r}-\gamma} \] Where \[ \gamma=1-\frac{v_0^2}{v_e^2} \] To find the travel time, we integrate: \[ T=2\int_{R}^{R/\gamma}\frac{dr}{v(r)}=\frac{2}{v_e}\int_{R}^{R/\gamma}\frac{dr}{\sqrt{\frac{R}{r}-\gamma}}=\frac{2R}{v_e}\int_{\gamma}^{1}\frac{du}{u^2\sqrt{u-\gamma}} \] \[ T=\frac{2R}{v_e}\frac{\tan^{-1}\left ( \sqrt{\frac{1}{\gamma}-1} \right )+\sqrt{\gamma-\gamma^2}}{\gamma^{3/2}}=\frac{2R}{v_e}\frac{\sin^{-1}(u)+u\sqrt{1-u^2}}{(1-u^2)^{3/2}} \] Where \(u=v_0/v_e\).
Optimal Path through a Planet
We want to find the best path through a planet of radius R, connecting two points \(2\alpha\) radians apart (great circle angle). We assume the planet is of uniform density. As is well known, the acceleration due to gravity a radius r from the center of the planet is given by: \[ a=-gr/R \] Where \(g\) is the surface gravitational acceleration. Thus, if it falls from the surface along a path through the planet, its speed at a distance r from the center will be given by \[ \tfrac{1}{2}mv^2=\tfrac{1}{2}m\frac{g}{R}\left ( R^2-r^2 \right ) \] \[ v(r)=\sqrt{\frac{g}{R}} \sqrt{R^2-r^2} \] Let us suppose it falls along the path specified by the function \(r(\theta)\), where r is even and \(r(\pm\alpha)=R\). The total time is given by \[ T=2\int_{0}^{\alpha}\frac{d\ell}{v}=2\sqrt{\frac{R}{g}}\int_{0}^{\alpha}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}d\theta \] In order to obtain conditions for the optimal path, then, we use calculus of variations. The Lagrangian is \[ L(r,r',\theta)=\frac{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}} \] Using the Beltrami Identity, we find: \[ \frac{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}-\frac{r'^2}{{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}}=\frac{r^2}{{\sqrt{r^2+r'^2}}{\sqrt{R^2-r^2}}}=C \] Let \(1+ \tfrac{1}{C^2}=1/q^2\). Rearranging, we find: \[ r'=r\sqrt{\frac{\left (1+ \tfrac{1}{C^2} \right )r^2-R^2}{R^2-r^2}}=\frac{r}{q}\sqrt{\frac{r^2-R^2q^2}{R^2-r^2}} \] As \(r'(0)=0\), this implies that \[ r(0)=Rq \] \[ r=\frac{r(0)}{R} \] Let us make the change of variables: \(u=r^2/R^2\). This then gives: \[ u'=2u\sqrt{\frac{\tfrac{1}{q^2}u-1}{1-u}} \] \[ u(0)=q^2 \] In order to determine this value, we can integrate the differential equation: \[ \frac{1}{2u} \sqrt{\frac{1-u}{\tfrac{1}{q^2}u-1}}du=d\theta \] \[ \int_{q^2}^{1}\frac{1}{2u} \sqrt{\frac{1-u}{\tfrac{1}{q^2}u-1}}du=\frac{\pi}{2}(1-q)=\int_{0}^{\alpha}d\theta=\alpha \] Thus \[ q=1-\frac{2\alpha}{\pi} \] We can then find the total travel time: \[ T=2\sqrt{\frac{R}{g}}\int_{0}^{\alpha}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}d\theta=2\sqrt{\frac{R}{g}}\int_{Rq}^{R}\frac{\sqrt{r^2(\theta)+r'^2(\theta)}}{\sqrt{R^2-r^2(\theta)}}\frac{1}{r'}dr \] \[ T=2\sqrt{\frac{R}{g}}\int_{Rq}^{R} \frac{q}{r} \sqrt{r^2+\frac{r^2}{q^2}{\frac{r^2-R^2q^2}{R^2-r^2}}}\frac{dr}{\sqrt{r^2-R^2q^2}} \] \[ T=\sqrt{\frac{R}{g}}\sqrt{1-q^2}\int_{Rq}^{R} \frac{2rdr}{\sqrt{R^2-r^2} \sqrt{r^2-R^2q^2}} \] \[ T=\sqrt{\frac{R}{g}}\sqrt{1-q^2}\int_{q^2}^{1} \frac{dx}{\sqrt{1-x^2} \sqrt{x^2-q^2}} \] \[ T=\pi \sqrt{\frac{R}{g}}\sqrt{1-q^2}=2\sqrt{\frac{R}{g}}\sqrt{\pi\alpha-\alpha^2} \] Below we show several trajectories along the optimal path for several values of alpha:
In fact, these solutions are hypocycloids.
